Your FREE AP Precalculus Practice Test 2026 – 240+ Q&A

Polynomial and Rational Functions (87)

1. A polynomial function f has a local maximum at $x = -2$ and a local minimum at $x = 5$, with no other extrema. What is the minimum possible degree of f?

   • A.Degree 2
   • B.Degree 3
   • C.Degree 5
   • D.Degree 4
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   Correct answer: Degree 3

   Degree 3 is correct. Two turning points require at least degree 3, because a degree-n polynomial has at most $n - 1$ turning points, and $3 - 1 = 2$. A quadratic allows only one turning point, so degree 2 is too small.

2. Using the rational zeros test, which value is a possible rational zero of $f(x) = 2x^{3} + 3x - 4$?

   • A.$x = 3$
   • B.$x = 5$
   • C.$x = \frac{1}{3}$
   • D.$x = \frac{1}{2}$
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   Correct answer: $x = \frac{1}{2}$

   $x = \frac{1}{2}$ is correct. The rational zeros theorem says any rational zero is a factor of the constant term 4 over a factor of the leading coefficient 2, giving candidates like $\pm 1, 2, 4, \frac{1}{2}$. The value $\frac{1}{2}$ fits this list, while $\frac{1}{3}$ and 5 do not because 3 and 5 are not factors of 2 or 4.

3. When $f(x) = x^{3} - 4x^{2} + x + 6$ is divided by $(x - 3)$ using synthetic division, what is the remainder?

   • A.$-3$
   • B.9
   • C.0
   • D.6
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   Correct answer: 0

   The remainder is 0. Evaluating $f(3) = 27 - 36 + 3 + 6 = 0$, and by the remainder theorem this equals the synthetic division remainder. A remainder of zero also confirms that $(x - 3)$ is a factor.

4. A rational function $r(x) = \frac{x^{2} - 4}{x^{2} + 1}$ has how many vertical asymptotes?

   • A.Zero
   • B.Two
   • C.One
   • D.Three
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   Correct answer: Zero

   Zero is correct. Vertical asymptotes occur where the denominator equals zero for real inputs, but $x^{2} + 1 = 0$ has no real solutions since $x^{2}$ is never negative. Therefore the function has no vertical asymptotes even though the numerator has real zeros.

5. The function $f(x) = \frac{x^{2} + 2x - 3}{x - 1}$ can be simplified for $x \neq 1$. What is the simplified form?

   • A.$x - 1$
   • B.$x + 3$
   • C.$x - 3$
   • D.$x + 1$
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   Correct answer: $x + 3$

   $x + 3$ is correct. The numerator factors as $(x + 3)(x - 1)$, and canceling the common $(x - 1)$ leaves $x + 3$, valid everywhere except $x = 1$ where a hole occurs. Choosing the wrong factor pair would give an incorrect simplification.

6. For the rational function $f(x) = \frac{3x^{2} - 5}{x - 2}$, polynomial long division gives a slant asymptote. What is that line?

   • A.$y = 3x + 6$
   • B.$y = 3x$
   • C.$y = 3x - 6$
   • D.$y = x + 2$
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   Correct answer: $y = 3x + 6$

   $y = 3x + 6$ is correct. Dividing $3x^{2} - 5$ by $x - 2$ yields a quotient of $3x + 6$ with a nonzero remainder, and the slant asymptote is exactly that linear quotient. The remainder term shrinks to zero for large inputs, so the graph approaches this line.

7. What is the value of the polynomial $p(x) = x^{4} - 2x^{2}$ at $x = -1$?

   • A.3
   • B.1
   • C.$-3$
   • D.$-1$
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   Correct answer: $-1$

   $p(-1) = -1$. Substituting gives $(-1)^{4} - 2(-1)^{2} = 1 - 2 \times 1 = 1 - 2 = -1$. Both even powers make the negative input positive before the subtraction.

8. A rational function $f(x) = \frac{x + 4}{x^{2} - 16}$ has a hole rather than an asymptote at which input?

   • A.$x = -4$
   • B.$x = 16$
   • C.$x = 4$
   • D.$x = 0$
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   Correct answer: $x = -4$

   $x = -4$ is correct. The denominator factors as $(x - 4)(x + 4)$, so the $(x + 4)$ factor cancels with the numerator, leaving a removable discontinuity at $x = -4$. The factor $(x - 4)$ does not cancel, so $x = 4$ is a genuine vertical asymptote.

9. Descartes' rule of signs is applied to $f(x) = x^{3} - x^{2} + x - 1$. How many sign changes appear in $f(x)$?

   • A.Three
   • B.Two
   • C.Zero
   • D.One
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   Correct answer: Three

   Three is correct. Reading the coefficient signs of $x^{3} - x^{2} + x - 1$ gives plus, minus, plus, minus, which switches sign three times. This count gives the maximum number of positive real zeros under Descartes' rule.

10. The graph of a polynomial passes through $(0, 0)$, $(1, 0)$, and $(-1, 0)$, each a simple crossing. Which is the lowest-degree polynomial consistent with this?

    • A.$f(x) = x(x - 1)^{2}$
    • B.$f(x) = x(x - 1)(x + 1)$
    • C.$f(x) = x^{2}(x - 1)$
    • D.$f(x) = (x - 1)(x + 1)$
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    Correct answer: $f(x) = x(x - 1)(x + 1)$

    $f(x) = x(x - 1)(x + 1)$ is correct. Three simple crossings at 0, 1, and -1 require the three distinct factors $x$, $(x - 1)$, and $(x + 1)$, each to the first power for crossing behavior. Repeated factors would cause a bounce, and dropping a factor would miss a required zero.

11. For $f(x) = x^{2}$, what is the difference quotient $\frac{f(x + h) - f(x)}{h}$ simplified?

    • A.$2x + h$
    • B.$x + h$
    • C.$2xh$
    • D.$2x$
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    Correct answer: $2x + h$

    $2x + h$ is correct. Expanding $f(x + h) = x^{2} + 2xh + h^{2}$ and subtracting $x^{2}$ leaves $2xh + h^{2}$, and dividing by h gives $2x + h$. This expression represents the average rate of change over an interval of width h.

12. A rational function has the equation $f(x) = \frac{2x^{2}}{x^{2} - 1}$. What is its horizontal asymptote?

    • A.$y = 0$
    • B.There is none
    • C.$y = 1$
    • D.$y = 2$
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    Correct answer: $y = 2$

    $y = 2$ is correct. The numerator and denominator both have degree 2, so the horizontal asymptote is the ratio of leading coefficients, $\frac{2}{1} = 2$. Equal degrees rule out an asymptote of $y = 0$, which requires a smaller numerator degree.

13. If a polynomial f has $f(2) = 0$ and $f(-3) = 0$, which product must divide $f(x)$?

    • A.$(x + 2)(x - 3)$
    • B.$(x + 2)(x + 3)$
    • C.$(x - 2)(x + 3)$
    • D.$(x - 2)(x - 3)$
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    Correct answer: $(x - 2)(x + 3)$

    $(x - 2)(x + 3)$ is correct. By the factor theorem, a zero at $x = 2$ gives the factor $(x - 2)$ and a zero at $x = -3$ gives the factor $(x + 3)$, so their product divides f. The other choices place the zeros at the wrong values.

14. A polynomial model for the volume of an open box is $V(x) = x(10 - 2x)(8 - 2x)$, where x is the cutout length. What is the domain that makes physical sense?

    • A.$0 < x < 4$
    • B.$x > 5$
    • C.All real numbers
    • D.$0 < x < 8$
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    Correct answer: $0 < x < 4$

    $0 < x < 4$ is correct. The cutout x must be positive, and each side length $10 - 2x$ and $8 - 2x$ must stay positive, requiring $x < 5$ and $x < 4$ respectively. The tighter restriction $x < 4$ controls, so the meaningful domain is $0 < x < 4$.

15. What is the leading term of the product $(3x^{2})(x^{3} - 2x)$?

    • A.$-6x^{3}$
    • B.$x^{5}$
    • C.$3x^{5}$
    • D.$3x^{6}$
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    Correct answer: $3x^{5}$

    $3x^{5}$ is correct. Multiplying the highest-degree terms $3x^{2}$ and $x^{3}$ gives $3 \times x^{2 + 3} = 3x^{5}$, which dominates end behavior. The term $-6x^{3}$ is a lower-degree product, not the leading term.

16. The graph of a rational function flattens toward $y = 0$ on both ends but spikes to infinity near $x = 4$. Which description fits?

    • A.Slant asymptote and a hole at $x = 4$
    • B.Two horizontal asymptotes
    • C.No asymptotes at all
    • D.Horizontal asymptote $y = 0$ and a vertical asymptote at $x = 4$
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    Correct answer: Horizontal asymptote $y = 0$ and a vertical asymptote at $x = 4$

    Horizontal asymptote $y = 0$ with a vertical asymptote at $x = 4$ is correct. Flattening toward zero on both ends indicates the horizontal asymptote $y = 0$, and the unbounded spike near $x = 4$ marks a vertical asymptote there. A hole would not cause the output to spike to infinity.

17. A degree 4 polynomial with positive leading coefficient has exactly two real zeros, both simple. How many times does its graph touch but not cross the x-axis?

    • A.Two times
    • B.One time
    • C.Zero times
    • D.Four times
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    Correct answer: Zero times

    Zero times is correct. Both real zeros are simple, meaning odd multiplicity, so the graph crosses at each rather than touching and turning. Touching without crossing requires even multiplicity, which is not present here.

18. When two polynomials of degrees 3 and 2 are added, what is the degree of the resulting polynomial?

    • A.2
    • B.3
    • C.5
    • D.6
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    Correct answer: 3

    Degree 3 is correct. Adding polynomials keeps the highest degree present, and since 3 is larger than 2, the sum has degree 3 as long as the degree-3 term is not canceled. Degrees are multiplied, not added, only when multiplying polynomials.

19. A rational function $f(x) = \frac{x^{3}}{x^{2} + 1}$ has what type of end behavior asymptote?

    • A.A horizontal asymptote $y = 0$
    • B.A horizontal asymptote $y = 1$
    • C.A slant asymptote $y = x$
    • D.No asymptote
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    Correct answer: A slant asymptote $y = x$

    A slant asymptote $y = x$ is correct. The numerator degree 3 is exactly one more than the denominator degree 2, so division gives a linear quotient of $y = x$ that the graph approaches. A horizontal asymptote would require the degrees to be equal or the numerator degree to be smaller.

20. The polynomial $f(x) = 6x^{2} - 13x + 6$ factors completely as which product?

    • A.$(2x - 3)(3x - 2)$
    • B.$(2x - 3)(3x + 2)$
    • C.$(6x - 1)(x - 6)$
    • D.$(2x + 3)(3x + 2)$
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    Correct answer: $(2x - 3)(3x - 2)$

    $(2x - 3)(3x - 2)$ is correct. Expanding gives $6x^{2} - 4x - 9x + 6 = 6x^{2} - 13x + 6$, matching the original. Both constants are negative to produce the positive constant 6 and negative middle term.

21. A polynomial $p(x)$ is divided by $(x + 2)$ and leaves a nonzero remainder of $-5$. What is $p(-2)$?

    • A.5
    • B.0
    • C.2
    • D.$-5$
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    Correct answer: $-5$

    $p(-2) = -5$. The remainder theorem states the remainder upon division by $(x + 2)$, which is $(x - (-2))$, equals $p(-2)$. Since that remainder is $-5$, $p(-2) = -5$, and a nonzero remainder confirms $(x + 2)$ is not a factor.

22. A polynomial function is even and has a zero at $x = 3$. Which other value must also be a zero?

    • A.$x = 0$
    • B.$x = -3$
    • C.$x = \frac{1}{3}$
    • D.$x = 6$
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    Correct answer: $x = -3$

    $x = -3$ is correct. An even function satisfies $f(-x) = f(x)$, so $f(-3) = f(3) = 0$, forcing a matching zero at $x = -3$. The symmetry of even functions pairs zeros at opposite inputs.

23. For the function $f(x) = \frac{5}{x^{2} + 4x + 4}$, what is the location and nature of the discontinuity?

    • A.A hole at $x = -2$
    • B.A vertical asymptote at $x = -2$
    • C.A vertical asymptote at $x = 2$
    • D.No discontinuity
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    Correct answer: A vertical asymptote at $x = -2$

    A vertical asymptote at $x = -2$ is correct. The denominator factors as $(x + 2)^{2}$, which is zero at $x = -2$, and since the numerator 5 is nonzero there, the function blows up to infinity, giving a vertical asymptote. There is no canceling factor, so it is not a hole.

24. A polynomial of degree 7 with real coefficients has 3 distinct non-real zeros listed. Why is this impossible?

    • A.All zeros of a degree 7 polynomial are real
    • B.A polynomial cannot have distinct zeros
    • C.Degree 7 polynomials have no complex zeros
    • D.Non-real zeros must come in conjugate pairs, so the count must be even
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    Correct answer: Non-real zeros must come in conjugate pairs, so the count must be even

    Conjugate pairs require an even count, which is correct. For real-coefficient polynomials, each non-real zero is paired with its conjugate, so non-real zeros always total an even number. Three is odd, making the listed situation impossible.

25. A polynomial passes through the point $(0, -8)$ and has no other information given. What is its constant term?

    • A.$-8$
    • B.Cannot be determined
    • C.0
    • D.8
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    Correct answer: $-8$

    The constant term is $-8$. The y-intercept of a polynomial equals its constant term, found by evaluating at $x = 0$, and the point $(0, -8)$ gives $f(0) = -8$ directly. So the constant term must be $-8$.

26. The graph of $f(x) = x^{3} - x$ is rewritten in factored form. Which factorization is correct?

    • A.$x(x^{2} + 1)$
    • B.$x^{2}(x - 1)$
    • C.$(x - 1)(x + 1)$
    • D.$x(x - 1)(x + 1)$
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    Correct answer: $x(x - 1)(x + 1)$

    $x(x - 1)(x + 1)$ is correct. Factoring out x gives $x(x^{2} - 1)$, and the difference of squares $x^{2} - 1$ factors into $(x - 1)(x + 1)$. This reveals three simple real zeros at 0, 1, and -1.

27. A rational function model for drug concentration is $C(t) = \frac{50t}{t^{2} + 9}$. As t increases without bound, what does C approach?

    • A.9
    • B.0
    • C.Infinity
    • D.50
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    Correct answer: 0

    C approaches 0. The denominator degree 2 exceeds the numerator degree 1, so the horizontal asymptote is $y = 0$ and concentration tends toward zero for large times. This models a drug clearing from the system over time.

28. What is the standard form expansion of $(x - 4)(x + 1)$?

    • A.$x^{2} + 3x - 4$
    • B.$x^{2} - 5x - 4$
    • C.$x^{2} - 3x - 4$
    • D.$x^{2} - 4x + 1$
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    Correct answer: $x^{2} - 3x - 4$

    $x^{2} - 3x - 4$ is correct. Multiplying gives $x^{2} + x - 4x - 4$, and combining the middle terms yields $x^{2} - 3x - 4$. The constant $-4$ comes from multiplying $-4$ by 1.

29. A function f is concave down on $(-\infty, 1)$ and concave up on $(1, \infty)$. What occurs at $x = 1$?

    • A.A point of inflection
    • B.A removable discontinuity
    • C.A vertical asymptote
    • D.A relative maximum
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    Correct answer: A point of inflection

    A point of inflection is correct. Concavity changing from down to up at $x = 1$ defines an inflection point, where the bending direction of the graph reverses. This is unrelated to extrema or to asymptotes, which involve direction changes or undefined behavior instead.

30. A rational function $r(x) = \frac{x - 3}{x^{2} - 9}$ is simplified. What is the simplified form and excluded value besides $x = -3$?

    • A.$\frac{1}{x - 3}$, with $x \neq 3$
    • B.$x - 3$, with no restriction
    • C.$\frac{1}{x + 3}$, with $x \neq 3$
    • D.$(x + 3)$, with $x \neq 3$
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    Correct answer: $\frac{1}{x + 3}$, with $x \neq 3$

    $\frac{1}{x + 3}$ with $x \neq 3$ is correct. The denominator factors as $(x - 3)(x + 3)$, and canceling the common $(x - 3)$ leaves $\frac{1}{x + 3}$, creating a hole at $x = 3$. The remaining factor $(x + 3)$ still gives a vertical asymptote at $x = -3$.

31. A degree 3 polynomial has end behavior falling on the left and rising on the right. What is true about its leading coefficient?

    • A.It cannot be determined from end behavior
    • B.It is zero
    • C.It is negative
    • D.It is positive
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    Correct answer: It is positive

    Positive is correct. An odd-degree polynomial that falls on the left and rises on the right matches a positive leading coefficient, the same pattern as the basic cubic. A negative leading coefficient would reverse the ends to rise on the left and fall on the right.

32. Which polynomial has exactly the zeros $x = 0$ (multiplicity 1), $x = 2$ (multiplicity 2), and degree 3?

    • A.$f(x) = x(x - 2)$
    • B.$f(x) = x(x - 2)^{2}$
    • C.$f(x) = x(x + 2)^{2}$
    • D.$f(x) = x^{2}(x - 2)$
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    Correct answer: $f(x) = x(x - 2)^{2}$

    $f(x) = x(x - 2)^{2}$ is correct. The factor x gives a simple zero at 0, and $(x - 2)^{2}$ gives a double zero at 2, and the total degree $1 + 2 = 3$ matches. The other forms misplace the multiplicities or the zero locations.

33. The average rate of change of $f(x) = x^{3}$ over the interval from $x = 0$ to $x = 2$ is what value?

    • A.8
    • B.2
    • C.4
    • D.6
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    Correct answer: 4

    The answer is 4. The average rate of change is $\frac{f(2) - f(0)}{2 - 0} = \frac{8 - 0}{2} = 4$. Forgetting to divide by the interval width of 2 would incorrectly give 8.

34. A rational function has numerator $(x^{2} + 1)$, which has no real zeros, and denominator $(x - 2)$. What does the lack of real numerator zeros tell you about the graph?

    • A.The graph has no x-intercepts
    • B.The graph has a hole at $x = 2$
    • C.The graph has two x-intercepts
    • D.The graph has no vertical asymptote
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    Correct answer: The graph has no x-intercepts

    No x-intercepts is correct. An x-intercept requires the numerator to be zero for a real input, but $x^{2} + 1$ is never zero for real x, so the graph never crosses the x-axis. The denominator zero at $x = 2$ still produces a vertical asymptote.

35. When dividing $f(x) = 2x^{2} + 3x - 2$ by $(x + 2)$, what is the quotient?

    • A.$2x + 1$
    • B.$x - 1$
    • C.$2x - 1$
    • D.$2x - 2$
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    Correct answer: $2x - 1$

    $2x - 1$ is correct. Since $f(-2) = 8 - 6 - 2 = 0$, $(x + 2)$ is a factor, and dividing gives the quotient $2x - 1$ because $(x + 2)(2x - 1) = 2x^{2} + 3x - 2$. The product check confirms the quotient.

36. A polynomial model of profit $P(x) = -2(x - 30)(x - 70)$ describes profit by number of units x. At which x values is profit zero?

    • A.$x = 0$ and $x = 100$
    • B.$x = -30$ and $x = -70$
    • C.$x = 2$ and $x = 50$
    • D.$x = 30$ and $x = 70$
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    Correct answer: $x = 30$ and $x = 70$

    $x = 30$ and $x = 70$ is correct. Setting each factor equal to zero gives $x = 30$ and $x = 70$, the break-even points where profit equals zero. The leading factor $-2$ affects the height of the parabola but not the locations of these zeros.

37. The reciprocal-squared function $f(x) = \frac{1}{x^{2}}$ has what end behavior and sign?

    • A.It approaches $y = 1$ on both ends
    • B.It approaches $y = 0$ from above on both ends
    • C.It grows without bound on both ends
    • D.It approaches $y = 0$ from below on both ends
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    Correct answer: It approaches $y = 0$ from above on both ends

    Approaching $y = 0$ from above is correct. Squaring keeps the denominator positive for all nonzero inputs, so outputs stay positive, and for large magnitude inputs $\frac{1}{x^{2}}$ shrinks toward zero. Thus the graph nears the horizontal asymptote $y = 0$ from above on both sides.

38. A polynomial f has a graph that crosses the x-axis at $x = -2$, bounces at $x = 1$, and crosses at $x = 4$. What is the smallest possible degree?

    • A.5
    • B.4
    • C.2
    • D.3
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    Correct answer: 4

    Degree 4 is correct. Crossing at -2 and 4 each requires odd multiplicity (at least 1 each), and bouncing at 1 requires even multiplicity (at least 2), giving a minimum total of $1 + 2 + 1 = 4$. A lower degree could not provide the bounce plus two crossings.

39. A polynomial $g(x) = (x^{2} + 5)$ has what kind of real zeros?

    • A.One real zero
    • B.Three real zeros
    • C.Two real zeros
    • D.No real zeros
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    Correct answer: No real zeros

    No real zeros is correct. Setting $x^{2} + 5 = 0$ gives $x^{2} = -5$, which has no real solution since a square is never negative. The zeros of this expression are non-real complex numbers, so the graph never touches the x-axis.

40. A polynomial f has zeros at $x = -2$, $x = 0$, and $x = 3$, each of multiplicity 1, and a positive leading coefficient with degree 3. What is the sign of $f(x)$ on the interval between $x = 0$ and $x = 3$?

    • A.Positive throughout the interval
    • B.Negative throughout the interval
    • C.Zero throughout the interval
    • D.Positive then negative within the interval
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    Correct answer: Negative throughout the interval

    Negative throughout is correct. With simple zeros at -2, 0, and 3 and a positive leading coefficient, the cubic is positive for $x > 3$, then alternates sign at each crossing, making it negative on $(0, 3)$. A single-multiplicity zero produces one sign change, so the function does not switch sign again inside the interval.

41. A polynomial $g(x)$ is the product $(x - 1)(x + 4)(x - 1)$. Which statement correctly describes its behavior at $x = 1$?

    • A.The graph crosses the x-axis steeply at $x = 1$
    • B.The graph touches the x-axis at $x = 1$ and turns back
    • C.The graph has a vertical asymptote at $x = 1$
    • D.The graph has a hole at $x = 1$
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    Correct answer: The graph touches the x-axis at $x = 1$ and turns back

    Touching and turning is correct. The factor $(x - 1)$ appears twice, giving $x = 1$ a multiplicity of 2, which is even, so the graph touches the axis there and reverses without crossing. Holes and asymptotes are rational-function features, not polynomial-zero behavior.

42. For the rational function $f(x) = \frac{x + 3}{x^{2} + x - 6}$, at which input is there a removable discontinuity?

    • A.$x = 2$
    • B.$x = -3$
    • C.$x = -2$
    • D.$x = 6$
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    Correct answer: $x = -3$

    $x = -3$ is correct. The denominator factors as $(x + 3)(x - 2)$, so the numerator $(x + 3)$ cancels with one denominator factor, removing the discontinuity at $x = -3$ and leaving a hole there. The remaining $(x - 2)$ factor does not cancel, so $x = 2$ is a vertical asymptote, not a hole.

43. Which polynomial has end behavior in which both ends rise and exactly two distinct real zeros?

    • A.$f(x) = -(x - 1)(x + 1)$
    • B.$f(x) = (x - 1)^{2} (x + 1)^{2}$
    • C.$f(x) = (x - 1)(x + 1)(x - 2)$
    • D.$f(x) = -(x - 1)^{2} (x + 1)^{2}$
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    Correct answer: $f(x) = (x - 1)^{2} (x + 1)^{2}$

    The fourth-degree product $(x - 1)^{2} (x + 1)^{2}$ is correct. Its degree is 4 (even) with a positive leading coefficient, so both ends rise, and it has exactly two distinct real zeros at 1 and -1 (each multiplicity 2). The negative versions send both ends down, and the cubic has opposite end behaviors.

44. A rational function $h(x) = \frac{x^{2} - 4}{x + 2}$ is simplified. What is the resulting expression and its restriction?

    • A.$x - 2$, with $x \neq -2$
    • B.$x + 2$, with $x \neq 2$
    • C.$x - 2$, with no restriction
    • D.$x + 2$, with $x \neq -2$
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    Correct answer: $x - 2$, with $x \neq -2$

    $x - 2$ with $x \neq -2$ is correct. The numerator is a difference of squares, $(x - 2)(x + 2)$; canceling the common $(x + 2)$ leaves $x - 2$. The original function is undefined at $x = -2$, so that input is excluded, leaving a hole there.

45. A polynomial of degree 4 has a positive leading coefficient. What is the maximum number of times its graph can intersect a horizontal line?

    • A.2
    • B.4
    • C.3
    • D.5
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    Correct answer: 4

    Four is correct. Setting the polynomial equal to a constant gives another degree-4 equation, which can have at most 4 real solutions, so the graph can meet any horizontal line at most 4 times. The count cannot exceed the degree.

46. Given $f(x) = 3x^{2}$ and the table values $f(1) = 3$, $f(2) = 12$, $f(3) = 27$, what are the first differences of the outputs over these equally spaced inputs?

    • A.3 and 12
    • B.9 and 15
    • C.6 and 6
    • D.12 and 27
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    Correct answer: 9 and 15

    9 and 15 is correct. The first differences subtract consecutive outputs: $12 - 3 = 9$ and $27 - 12 = 15$. These are not constant, which is expected for a quadratic, whose second differences $(15 - 9 = 6)$ are the constant ones instead.

47. A rational function $r(x)$ has a numerator of degree 2 and a denominator of degree 5. Which describes its end behavior?

    • A.It approaches a slant asymptote
    • B.It increases without bound on both ends
    • C.It approaches the horizontal asymptote $y = 0$
    • D.It has no asymptotes
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    Correct answer: It approaches the horizontal asymptote $y = 0$

    Approaching $y = 0$ is correct. Because the denominator degree (5) exceeds the numerator degree (2), the denominator dominates for large inputs and the outputs shrink toward zero. A slant asymptote would require the numerator degree to exceed the denominator degree by exactly one.

48. A polynomial $p(x)$ yields $p(-1) = 0$. Which factor must appear in its factored form?

    • A.$(x - 1)$
    • B.$(x + 1)$
    • C.$(x + 0)$
    • D.$(1 - x)$
    Show answerHide answer

    Correct answer: $(x + 1)$

    $(x + 1)$ is correct. By the Factor Theorem, a zero at $x = c$ corresponds to the factor $(x - c)$; here $c = -1$, so the factor is $(x - (-1)) = (x + 1)$. The factor $(x - 1)$ would correspond to a zero at $x = 1$ instead.

49. The graph of a function rises to a peak, falls to a valley, rises to another peak, and then falls. What is the minimum polynomial degree consistent with this shape?

    • A.Degree 3
    • B.Degree 4
    • C.Degree 2
    • D.Degree 5
    Show answerHide answer

    Correct answer: Degree 4

    Degree 4 is correct. The described shape has three turning points (peak, valley, peak), and a degree-n polynomial allows at most $n - 1$ turning points, so three turning points require at least degree 4. A cubic permits only two turning points.

50. Which expression is a fully factored polynomial with a zero of multiplicity 3 at $x = 4$ and no other real zeros?

    • A.$(x - 4)^{3}$
    • B.$(x + 4)^{3}$
    • C.$(x - 4)^{2}$
    • D.$(x - 4)(x^{2} + 1)$
    Show answerHide answer

    Correct answer: $(x - 4)^{3}$

    $(x - 4)^{3}$ is correct. The factor $(x - 4)$ raised to the third power gives a single real zero at $x = 4$ with multiplicity 3 and no other real zeros. The version with $(x + 4)$ places the zero at -4, and the squared form has multiplicity 2.

51. For $f(x) = \frac{x - 3}{x^{2} - 9}$, what feature occurs at $x = 3$?

    • A.A vertical asymptote
    • B.An x-intercept
    • C.A horizontal asymptote
    • D.A hole
    Show answerHide answer

    Correct answer: A hole

    A hole is correct. The denominator factors as $(x - 3)(x + 3)$, and the $(x - 3)$ in the numerator cancels with one in the denominator, removing the discontinuity at $x = 3$ and leaving a hole. The remaining $(x + 3)$ factor produces a vertical asymptote at $x = -3$ instead.

52. A polynomial model $f(t)$ gives the height of a drone in meters at time $t$ seconds. If $f$ has a relative minimum at $(6, 10)$, what does this mean in context?

    • A.The drone is highest at 6 seconds
    • B.The drone reaches a local low of 10 meters at 6 seconds before rising again
    • C.The drone is on the ground at 6 seconds
    • D.The drone descends forever after 6 seconds
    Show answerHide answer

    Correct answer: The drone reaches a local low of 10 meters at 6 seconds before rising again

    A local low of 10 meters at 6 seconds is correct. A relative minimum is a point where the output is smaller than at nearby inputs, so the drone dips to 10 meters at 6 seconds and then climbs. It is not the overall lowest point nor ground level unless the height there were zero.

53. Which statement about the graph of an odd-degree polynomial with a negative leading coefficient is correct?

    • A.The left end falls and the right end rises
    • B.Both ends fall
    • C.Both ends rise
    • D.The left end rises and the right end falls
    Show answerHide answer

    Correct answer: The left end rises and the right end falls

    Left rises and right falls is correct. Odd degree gives opposite end behaviors, and a negative leading coefficient makes the right end fall while the left end rises. Both ends going the same way would require an even degree.

54. A rational function $f(x) = \frac{4x^{2} - 1}{2x^{2} + 3}$ has what horizontal asymptote?

    • A.$y = 4$
    • B.$y = 0$
    • C.$y = 2$
    • D.There is no horizontal asymptote
    Show answerHide answer

    Correct answer: $y = 2$

    $y = 2$ is correct. The numerator and denominator both have degree 2, so the horizontal asymptote is the ratio of leading coefficients, $\frac{4}{2} = 2$. A value of $y = 0$ would require the numerator degree to be smaller than the denominator degree.

55. If a polynomial has the factored form $f(x) = (x - 5)(x + 2)^{3}$, what is the total degree of $f$?

    • A.5
    • B.3
    • C.4
    • D.2
    Show answerHide answer

    Correct answer: 4

    Degree 4 is correct. The total degree is the sum of the exponents on the factors: $(x - 5)$ contributes 1 and $(x + 2)^{3}$ contributes 3, giving $1 + 3 = 4$. The degree counts multiplicities, not just the number of distinct factors.

56. A polynomial passes through $(0, -6)$ and has the form $f(x) = a(x - 1)(x + 3)$. What is the value of $a$?

    • A.2
    • B.-2
    • C.6
    • D.-6
    Show answerHide answer

    Correct answer: 2

    $a = 2$ is correct. Substituting $x = 0$ gives $f(0) = a(0 - 1)(0 + 3) = a \times (-1) \times 3 = -3a$, and setting this equal to -6 yields $-3a = -6$, so $a = 2$. The other values do not satisfy the point.

57. The graph of a rational function has vertical asymptotes at $x = -1$ and $x = 4$. Which denominator is consistent with this?

    • A.$(x - 1)(x + 4)$
    • B.$(x + 1)(x - 4)$
    • C.$(x + 1)(x + 4)$
    • D.$(x - 1)(x - 4)$
    Show answerHide answer

    Correct answer: $(x + 1)(x - 4)$

    $(x + 1)(x - 4)$ is correct. Vertical asymptotes occur where the denominator is zero and does not cancel; setting $(x + 1) = 0$ gives $x = -1$ and $(x - 4) = 0$ gives $x = 4$, matching the asymptotes. The other denominators place the zeros at the wrong values.

58. A polynomial $f$ satisfies $f(2) = 5$ and $f(6) = 13$. What is its average rate of change on the interval from $x = 2$ to $x = 6$?

    • A.8
    • B.4
    • C.2
    • D.3
    Show answerHide answer

    Correct answer: 2

    The average rate of change is 2. It equals $\frac{f(6) - f(2)}{6 - 2} = \frac{13 - 5}{4} = \frac{8}{4} = 2$. Forgetting to divide by the interval width of 4 would give the incorrect value of 8.

59. Which describes the relationship between the factors of a polynomial and the x-intercepts where the graph crosses versus touches the axis?

    • A.Even-multiplicity factors cause crossing; odd-multiplicity factors cause touching
    • B.Odd-multiplicity factors cause crossing; even-multiplicity factors cause touching
    • C.All factors cause the graph to cross
    • D.All factors cause the graph to touch and turn
    Show answerHide answer

    Correct answer: Odd-multiplicity factors cause crossing; even-multiplicity factors cause touching

    Odd causes crossing and even causes touching is correct. A factor with odd multiplicity makes the graph pass through the axis, while a factor with even multiplicity makes it touch and turn back. The behavior depends on whether the multiplicity is odd or even.

60. A rational function $f(x) = \frac{x^{2} + 1}{x - 2}$ has numerator degree one more than denominator degree. What type of asymptote does it have besides the vertical one?

    • A.A horizontal asymptote at $y = 1$
    • B.A slant asymptote
    • C.A second vertical asymptote
    • D.No additional asymptote
    Show answerHide answer

    Correct answer: A slant asymptote

    A slant asymptote is correct. Since the numerator degree (2) exceeds the denominator degree (1) by exactly one, dividing gives a linear quotient that the graph approaches, producing a slant asymptote. There is no horizontal asymptote when the numerator degree is larger.

61. If a polynomial of degree 7 has 5 distinct real zeros and one of them has multiplicity 3 while the rest are simple, how many non-real zeros does it have, counted with multiplicity?

    • A.0
    • B.2
    • C.4
    • D.1
    Show answerHide answer

    Correct answer: 0

    Zero non-real zeros is correct. The real zeros account for 3 (from the triple) plus 4 simple zeros $= 7$ total with multiplicity, which equals the degree. Since all 7 zeros are real, none are non-real.

62. The function $f(x) = x^{3}$ is transformed into $g(x) = \frac{1}{2} \times x^{3}$. How does this affect the graph?

    • A.A vertical compression by a factor of $\frac{1}{2}$
    • B.A horizontal stretch by a factor of $\frac{1}{2}$
    • C.A vertical stretch by a factor of 2
    • D.A reflection over the x-axis
    Show answerHide answer

    Correct answer: A vertical compression by a factor of $\frac{1}{2}$

    A vertical compression by $\frac{1}{2}$ is correct. Multiplying the output by a constant between 0 and 1 shrinks every y-value toward the x-axis, compressing the graph vertically. A factor greater than 1 would stretch it instead, and a negative factor would add a reflection.

63. A polynomial model for a company's monthly profit has a y-intercept of -8 (in thousands of dollars). What does this represent in context?

    • A.The break-even production level
    • B.The maximum monthly profit
    • C.The profit when no units are produced, an 8 thousand dollar loss
    • D.The number of units that maximize profit
    Show answerHide answer

    Correct answer: The profit when no units are produced, an 8 thousand dollar loss

    An 8 thousand dollar loss at zero production is correct. The y-intercept is the output when the input is zero, so it gives the profit when no units are made, here a loss of 8 thousand dollars from fixed costs. Break-even points are zeros, not the y-intercept.

64. Which polynomial is written in standard form?

    • A.$f(x) = 3 + 2x^{2} - x$
    • B.$f(x) = -x + 3 + 2x^{2}$
    • C.$f(x) = 2x^{2} - x + 3$
    • D.$f(x) = 2x^{2} + 3 - x$
    Show answerHide answer

    Correct answer: $f(x) = 2x^{2} - x + 3$

    $2x^{2} - x + 3$ is correct. Standard form lists terms in order of decreasing exponent, here the squared term, then the linear term, then the constant. The other expressions contain the same terms but are not arranged from highest to lowest degree.

65. A rational function $f(x) = \frac{6}{x^{2} + 2}$ has what kind of vertical asymptotes?

    • A.One vertical asymptote at $x = 0$
    • B.No vertical asymptotes
    • C.Two vertical asymptotes
    • D.A vertical asymptote at $x = 2$
    Show answerHide answer

    Correct answer: No vertical asymptotes

    No vertical asymptotes is correct. The denominator $x^{2} + 2$ is always at least 2 and never equals zero for any real input, so the function is defined everywhere and has no vertical asymptotes. Vertical asymptotes require real zeros of the denominator.

66. Two equally spaced data sets are compared. Set A has constant first differences and Set B has constant second differences. Which polynomial degrees do they suggest?

    • A.Set A linear, Set B quadratic
    • B.Set A quadratic, Set B linear
    • C.Set A cubic, Set B quadratic
    • D.Both linear
    Show answerHide answer

    Correct answer: Set A linear, Set B quadratic

    Set A linear and Set B quadratic is correct. Constant first differences indicate a degree-1 (linear) pattern, and constant second differences indicate a degree-2 (quadratic) pattern. The order of differences that becomes constant matches the polynomial degree.

67. The polynomial $f(x) = -(x + 1)(x - 2)(x - 5)$ is evaluated for very large negative $x$. What is the end behavior on the left?

    • A.$f(x)$ rises toward positive infinity
    • B.$f(x)$ approaches a horizontal asymptote
    • C.$f(x)$ approaches zero
    • D.$f(x)$ falls toward negative infinity
    Show answerHide answer

    Correct answer: $f(x)$ rises toward positive infinity

    Rising on the left is correct. Expanding gives a leading term of $-x^{3}$ (odd degree, negative leading coefficient), so the left end rises while the right end falls. For very large negative $x$, the negative coefficient times a large negative cube yields large positive outputs.

68. A polynomial graph touches the x-axis at $x = 0$, crosses at $x = 3$, and touches at $x = -4$. What is the minimum possible degree?

    • A.3
    • B.5
    • C.4
    • D.6
    Show answerHide answer

    Correct answer: 5

    Degree 5 is correct. Touching at $x = 0$ requires multiplicity at least 2, crossing at $x = 3$ requires multiplicity at least 1, and touching at $x = -4$ requires multiplicity at least 2, summing to $2 + 1 + 2 = 5$. This is the smallest total degree consistent with the described behavior.

69. Which transformation of $f(x) = x^{2}$ produces a parabola with vertex at $(3, -1)$ that opens upward?

    • A.$g(x) = (x - 3)^{2} - 1$
    • B.$g(x) = (x + 3)^{2} - 1$
    • C.$g(x) = (x - 3)^{2} + 1$
    • D.$g(x) = -(x - 3)^{2} - 1$
    Show answerHide answer

    Correct answer: $g(x) = (x - 3)^{2} - 1$

    $g(x) = (x - 3)^{2} - 1$ is correct. The $(x - 3)$ shifts the vertex right to $x = 3$, and the minus 1 shifts it down to $y = -1$, while the positive coefficient keeps it opening upward. The negative version would open downward, and $(x + 3)$ would move the vertex left.

70. A rational function $f(x) = \frac{3x}{x - 2}$ has what horizontal asymptote?

    • A.$y = 3$
    • B.$y = 0$
    • C.$y = 2$
    • D.There is no horizontal asymptote
    Show answerHide answer

    Correct answer: $y = 3$

    $y = 3$ is correct. The numerator and denominator both have degree 1, so the horizontal asymptote is the ratio of leading coefficients, $\frac{3}{1} = 3$. The function levels off toward $y = 3$ for large magnitude inputs.

71. If a polynomial has real coefficients and exactly the zeros 2, $2i$, and $-2i$, what is its minimum possible degree?

    • A.2
    • B.1
    • C.4
    • D.3
    Show answerHide answer

    Correct answer: 3

    Degree 3 is correct. The real zero 2 contributes one factor, and the non-real zeros $2i$ and $-2i$ form a conjugate pair contributing two more, giving a minimum of $1 + 2 = 3$. Conjugate pairs are required because the coefficients are real.

72. A polynomial $f$ is decreasing on the interval $(1, 5)$. What can be said about its average rate of change over that interval?

    • A.It is positive
    • B.It is zero
    • C.It is negative
    • D.It cannot be determined
    Show answerHide answer

    Correct answer: It is negative

    Negative is correct. If a function is decreasing across an entire interval, its output at the right endpoint is lower than at the left endpoint, so the net change is negative and the average rate of change is negative. A decreasing function cannot have a positive average rate of change over the interval.

73. Which polynomial has a graph that is symmetric about the origin?

    • A.$f(x) = x^{3} - 4x$
    • B.$f(x) = x^{2} + 1$
    • C.$f(x) = x^{3} + 2$
    • D.$f(x) = x^{4}$
    Show answerHide answer

    Correct answer: $f(x) = x^{3} - 4x$

    $f(x) = x^{3} - 4x$ is correct. Both terms have odd exponents and there is no constant term, so the function is odd and its graph is symmetric about the origin. The other options contain even-power terms or a nonzero constant, which break origin symmetry.

74. A rational function $f(x) = \frac{(x - 1)(x + 2)}{(x + 2)(x - 4)}$ is graphed. Which features appear at $x = -2$ and $x = 4$ respectively?

    • A.Vertical asymptote at -2 and hole at 4
    • B.Hole at -2 and vertical asymptote at 4
    • C.Holes at both -2 and 4
    • D.Vertical asymptotes at both -2 and 4
    Show answerHide answer

    Correct answer: Hole at -2 and vertical asymptote at 4

    Hole at -2 and vertical asymptote at 4 is correct. The factor $(x + 2)$ cancels between numerator and denominator, producing a hole at $x = -2$, while the non-canceling $(x - 4)$ leaves the denominator zero at $x = 4$, producing a vertical asymptote there.

75. A polynomial model for the volume of a box is $V(x) = x(10 - 2x)(8 - 2x)$, where $x$ is the height. For what positive inputs is the volume model physically meaningful (positive volume)?

    • A.$0 < x < 4$
    • B.$0 < x < 5$
    • C.$x > 5$
    • D.All positive $x$
    Show answerHide answer

    Correct answer: $0 < x < 4$

    $0 < x < 4$ is correct. Volume must be positive, which requires all three factors positive: $x > 0$, $10 - 2x > 0$ (so $x < 5$), and $8 - 2x > 0$ (so $x < 4$). The most restrictive bound gives $0 < x < 4$, where every dimension is positive.

76. What is the result of dividing the polynomial $x^{2} - x - 6$ by $(x - 3)$?

    • A.$x + 2$
    • B.$x - 2$
    • C.$x + 3$
    • D.$x - 6$
    Show answerHide answer

    Correct answer: $x + 2$

    $x + 2$ is correct. The dividend factors as $(x - 3)(x + 2)$, so dividing by $(x - 3)$ leaves the quotient $x + 2$ with no remainder. This also confirms that $x = 3$ is a zero of the original polynomial.

77. A polynomial of even degree has a positive leading coefficient and a negative value at its lowest turning point. How many real zeros must it have at minimum?

    • A.Exactly 1
    • B.At least 2
    • C.Exactly 0
    • D.Exactly the degree
    Show answerHide answer

    Correct answer: At least 2

    At least 2 is correct. With both ends rising (positive even-degree) and a turning point dipping below the x-axis, the graph must cross the axis on the way down and again on the way back up, giving at least two real zeros. An even-degree polynomial that dips below zero cannot have zero real zeros.

78. The reciprocal-squared function $f(x) = \frac{1}{x^{2}}$ differs from $f(x) = \frac{1}{x}$ in which way near its vertical asymptote at $x = 0$?

    • A.$\frac{1}{x^{2}}$ goes to positive infinity on both sides, while $\frac{1}{x}$ goes to opposite infinities
    • B.Both go to negative infinity on both sides
    • C.$\frac{1}{x^{2}}$ goes to opposite infinities on each side
    • D.Both behave identically near zero
    Show answerHide answer

    Correct answer: $\frac{1}{x^{2}}$ goes to positive infinity on both sides, while $\frac{1}{x}$ goes to opposite infinities

    $\frac{1}{x^{2}}$ rising on both sides is correct. Squaring the denominator keeps outputs positive near zero, so $\frac{1}{x^{2}}$ tends to positive infinity from both sides, whereas $\frac{1}{x}$ is positive on one side and negative on the other, giving opposite infinities. The even power changes the sign behavior near the asymptote.

79. A polynomial $p(x)$ has $p(0) = 4$ and is known to be even. Which point must also lie on its graph?

    • A.$(-4, 0)$
    • B.$(4, 0)$
    • C.There is not enough information
    • D.$(0, -4)$
    Show answerHide answer

    Correct answer: There is not enough information

    Not enough information is correct. Knowing $p(0) = 4$ fixes only the single point $(0, 4)$, and evenness gives symmetry about the y-axis but does not by itself determine the output at any other specific input. No additional point can be guaranteed from the y-intercept and symmetry alone.

80. Which expression is the complete factorization of $x^{3} - 8$?

    • A.$(x - 2)(x^{2} + 2x + 4)$
    • B.$(x - 2)(x^{2} - 2x + 4)$
    • C.$(x - 2)(x^{2} - 4)$
    • D.$(x + 2)(x^{2} - 2x + 4)$
    Show answerHide answer

    Correct answer: $(x - 2)(x^{2} + 2x + 4)$

    The factorization $(x - 2)(x^{2} + 2x + 4)$ is correct. As a difference of cubes, $x^{3} - 2^{3}$ factors into $(x - 2)$ times the quadratic $x^{2} + 2x + 4$, using the pattern $a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})$. The choice with a plus sign in the linear factor or the wrong sign in the quadratic does not match the difference-of-cubes pattern.

81. When the polynomial $p(x) = x^{3} - 2x^{2} + x - 5$ is divided by $(x - 2)$, what is the remainder?

    • A.5
    • B.0
    • C.-3
    • D.-5
    Show answerHide answer

    Correct answer: -3

    The remainder is -3. By the Remainder Theorem the remainder equals $p(2)$, and substituting gives 8 minus 8 plus 2 minus 5 = -3. Computing $p(2)$ directly avoids carrying out the full division, and the nonzero result confirms that $(x - 2)$ is not a factor.

82. A polynomial with real coefficients has degree 4 and zeros at $x = 1$, $x = 1$, and $x = 2i$. What is its remaining zero?

    • A.$x = 1$
    • B.$x = -2i$
    • C.$x = 2$
    • D.$x = -1$
    Show answerHide answer

    Correct answer: $x = -2i$

    The remaining zero is $x = -2i$. A degree-4 polynomial has four zeros counted with multiplicity, and three are accounted for by the double zero at $x = 1$ and the single zero at $x = 2i$. Because real coefficients force non-real zeros into conjugate pairs, the partner of $2i$ must be its conjugate $-2i$.

83. Algebraically, which test confirms that $f(x) = x^{3} - 4x$ is an odd function?

    • A.Showing $f(-x) = f(x)$ for all x
    • B.Showing $f(-x) = -f(x)$, since $f(-x) = -x^{3} + 4x = -(x^{3} - 4x)$
    • C.Showing every exponent is even
    • D.Showing $f(0) = 0$
    Show answerHide answer

    Correct answer: Showing $f(-x) = -f(x)$, since $f(-x) = -x^{3} + 4x = -(x^{3} - 4x)$

    Showing $f(-x) = -f(x)$ is correct. Substituting $-x$ gives $(-x)^{3}$ minus $4(-x) = -x^{3}$ plus 4x, which is exactly the opposite of the original $f(x) = x^{3} - 4x$, confirming the function is odd. The condition $f(-x) = f(x)$ would instead define an even function, and $f(0) = 0$ alone does not establish odd symmetry.

84. For $f(x) = x^{2} + x$, the average rate of change on $[0, 2]$ compared to the average rate of change on $[2, 4]$ is which of the following?

    • A.Smaller, because the function is concave up
    • B.Larger, because the function is concave up
    • C.Equal, because the function is a polynomial
    • D.Smaller, because the function is concave down
    Show answerHide answer

    Correct answer: Smaller, because the function is concave up

    Smaller is correct. On $[0, 2]$ the rate equals [6 minus 0] over 2 = 3, while on $[2, 4]$ it equals [20 minus 6] over 2 = 7, so the earlier interval has the smaller rate. This rising sequence of average rates of change is the hallmark of a concave up graph.

85. What are the coordinates of the vertex of the quadratic $f(x) = (x - 4)^{2} - 9$?

    • A.$(-4, -9)$
    • B.$(4, 9)$
    • C.$(4, -9)$
    • D.$(-4, 9)$
    Show answerHide answer

    Correct answer: $(4, -9)$

    The vertex is $(4, -9)$. In vertex form $a(x - h)^{2} + k$, the vertex is the point $(h, k)$; here $h = 4$ and $k = -9$. The subtraction inside the squared term means the horizontal coordinate is positive 4, not -4, which the incorrect choices reverse.

86. The graph of $g(x) = \frac{1}{x - 3} + 2$ is the reciprocal function $f(x) = \frac{1}{x}$ transformed how?

    • A.Shifted left 3 and down 2, with asymptotes $x = -3$ and $y = -2$
    • B.Shifted right 3 and up 2, with asymptotes $x = 3$ and $y = 2$
    • C.Shifted right 2 and up 3, with asymptotes $x = 2$ and $y = 3$
    • D.Shifted left 3 and up 2, with asymptotes $x = -3$ and $y = 2$
    Show answerHide answer

    Correct answer: Shifted right 3 and up 2, with asymptotes $x = 3$ and $y = 2$

    Right 3 and up 2 is correct. Replacing x with $x - 3$ moves the graph right 3, sending the vertical asymptote to $x = 3$, and adding 2 outside moves it up 2, sending the horizontal asymptote to $y = 2$. The inside subtraction shifts the graph in the positive x-direction, so a leftward shift does not apply.

87. In the factored polynomial $f(x) = (x + 1)(x - 2)(x - 2)(x + 1)$, what is the multiplicity of the zero at $x = 2$ and how does the graph behave there?

    • A.Multiplicity 1; the graph crosses the x-axis
    • B.Multiplicity 4; the graph crosses the x-axis
    • C.Multiplicity 3; the graph flattens and crosses
    • D.Multiplicity 2; the graph touches the x-axis and turns around
    Show answerHide answer

    Correct answer: Multiplicity 2; the graph touches the x-axis and turns around

    Multiplicity 2 with a touch-and-turn is correct. The factor $(x - 2)$ appears twice, giving the zero at $x = 2$ a multiplicity of 2, and an even multiplicity causes the graph to touch the x-axis and bounce back without crossing. The factor $(x + 1)$ also appears twice but governs the separate zero at $x = -1$.